Simplify the following expression and state the condition under which the simplification is valid. You can assume that $q \neq 0$. $a = \dfrac{3}{10q - 80} \times \dfrac{9(q - 8)}{3q} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ 3 \times 9(q - 8) } { (10q - 80) \times 3q } $ $ a = \dfrac {3 \times 9(q - 8)} {3q \times 10(q - 8)} $ $ a = \dfrac{27(q - 8)}{30q(q - 8)} $ We can cancel the $q - 8$ so long as $q - 8 \neq 0$ Therefore $q \neq 8$ $a = \dfrac{27 \cancel{(q - 8})}{30q \cancel{(q - 8)}} = \dfrac{27}{30q} = \dfrac{9}{10q} $